Question

# If one zero of the polynomial$$\left( { a }^{ 2 }+9 \right) { x }^{ 2 }+13x+6a$$ is reciprocal of the other, find the value of a

Solution

## $$\Rightarrow$$  The given quadratic polynomial is $$(a^2+9)x^2+13x+6a$$.$$\Rightarrow$$  Let one zero of the quadratic polynomial be $$\alpha$$.$$\therefore$$  The other zero of the quadratic polynomial is $$\dfrac{1}{\alpha}.$$$$\Rightarrow$$  $$\alpha\times \dfrac{1}{\alpha}=\dfrac{6a}{a^2+9}$$$$\Rightarrow$$  $$1=\dfrac{6a}{a^2+9}$$$$\Rightarrow$$  $$a^2+9=6a$$$$\Rightarrow$$  $$a^2-6a+9=0$$$$\Rightarrow$$  $$(a-3)^2=0$$$$\Rightarrow$$  $$a-3=0$$$$\Rightarrow$$  $$a=3$$$$\therefore$$  The value of $$a$$ is $$3$$.  Mathematics

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